Question

$Ifx=\sec A+\sin Aandy=\sec A-\sin A,$ prove that
${\left(\frac{2}{x+y}\right)}^2+{\left(\frac{x-y}{2}\right)}^2=1$

Answer 1

We have,

$x=\sec A+\sin A$ and $y=\sec A-\sin A$

Since,

${\left(\frac{2}{x+y}\right)}^2+{\left(\frac{x-y}{2}\right)}^2$

$\Rightarrow {\left(\frac{2}{\sec A+\sin A+\sec A-\sin A}\right)}^2+{\left(\frac{\sec A+\sin A-\sec A+\sin A}{2}\right)}^2$

$\Rightarrow {\left(\frac{2}{2\sec A}\right)}^2+{\left(\frac{2\sin A}{2}\right)}^2$

$\Rightarrow {\left(\frac{1}{\sec A}\right)}^2+{\left(\sin A\right)}^2$

$\Rightarrow {\cos }^{2}A+{\sin }^{2}A$

$\Rightarrow 1$

Hence, proved.